Question

我是swift的新手，并尝试做类似的事情：

我有一个名为Api Response的结构：

struct ApiResponse {
    var IsSuccess : Bool = false
    var Message : String?
    var ReturnedData : Data?
}

并在名为CommonHandler的另一个类中有一个func，它使api调用

public class CommonHandler {

    public func CallApi(_ apiUrl : String , _ parameters : [String : Any] ) -> ApiResponse
    {
        var apiResponse = ApiResponse()

        let url = URL(string: apiUrl)!
        var request = URLRequest(url: url)
        request.httpMethod = "POST"
        request.setValue("application/json", forHTTPHeaderField: "Content-Type")
        request.httpBody = try! JSONSerialization.data(withJSONObject: parameters, options: .prettyPrinted)

        let task = URLSession.shared.dataTask(with: request , completionHandler : { data, response, error in
            if let error = error {
                // handle the transport error
                return
            }
            guard let response = response as? HTTPURLResponse, response.statusCode == 200 else {
                // handle the server error
                return
            }
            apiResponse.ReturnedData = data
            apiResponse.IsSuccess = true
            apiResponse.Message = "Succeed"

        })
        task.resume()

        return apiResponse
    }
}

我想在UIViewController中调用这个函数：

var handler = CommonHandler()
let param :[String : String] = ["param":"param"]
let url = "url"

let response = handler.CallApi(url, param)
print(response.IsSuccess)
print(response.Message!)

我知道我不能像这样使用dataTask方法。这是异步的。但是我如何在非void函数中调用api并返回其响应数据？我将ReturnedData json解析为struct然后。在这种情况下，最好的方法是什么？感谢

Answer 1

使用异步完成处理程序

public func callApi(with url : String , parameters : [String : Any], completion: @escaping (ApiResponse?) -> () )
{
    var apiResponse = ApiResponse()

    let url = URL(string: apiUrl)!
    var request = URLRequest(url: url)
    request.httpMethod = "POST"
    request.setValue("application/json", forHTTPHeaderField: "Content-Type")
    request.httpBody = try! JSONSerialization.data(withJSONObject: parameters, options: .prettyPrinted)

    let task = URLSession.shared.dataTask(with: request , completionHandler : { data, response, error in
        if let error = error {
            // handle the transport error
            completion(nil)
            return
        }
        guard let response = response as? HTTPURLResponse, response.statusCode == 200 else {
            // handle the server error
            completion(nil)
            return
        }
        apiResponse.ReturnedData = data
        apiResponse.IsSuccess = true
        apiResponse.Message = "Succeed"
        completion(apiResponse)

    })
    task.resume()
}

并称之为：

var handler = CommonHandler()
let param = ["param":"param"]
let url = "url"

handler.callApi(with: url, parameters: param) { response in
    if let response = response {
       print(response.IsSuccess)
       print(response.Message!)
    }
}

注意：

请遵守变量和函数名称以小写字母开头的命名约定，并使用参数标签以提高可读性。

如何从包含DataTask的方法返回一个对象

1 个答案: